Chapter 4. Linear Second Order Equations
ay ′′ + by ′ + cy = 0,
(1)
where a, b, c are constants. The associated auxiliary equation is
ar 2 + br + c = 0.
(2)
Consequently, y = erx is a solution to (1) if an only if r satisfies (2).
So, the equation (2) is a quadratic, and its roots are:
√
√
−b + b2 − 4ac
−b − b2 − 4ac
r1 =
, r2 =
2a
2a
√
When b2 − 4ac > 0, then r1 , r2 ∈ R and r1 6= r2 . So, y1 (x) = er1 x and y2 (x) = er2 x are
two linearly independent solutions to (1) and
y(x) = c1 er1 x + c2 er2 x
is the general solution to (1).
√
b
If b2 − 4ac = 0, then the equation (2) has a repeated root r ∈ R, r = − 2a
. In this case,
rx
rx
y1 (x) = e and y2 (x) = xe are two linearly independent solutions to (1) and
y(x) = c1 erx + c2 xerx = (c1 + c2 x)erx
is the general solution to (1).
Example 1.
(a) Find the general solution to the equation
y ′′ − 2y ′ − 8y = 0.
SOLUTION
The associated auxiliary equation is
r 2 − 2r − 8 = 0,
which roots are r1 = 4, r2 = −2, the fundamental solution set is {e4x , e−2x }. Thus, the
general solution is
y(x) = c1 e4x + c2 e−2x .
(b) Solve the given initial value problem
4y ′′ − 12y ′ + 9y = 0, y(0) = 1, y ′(0) = 23 .
SOLUTION The associated characteristic equation is
4r 2 − 12r + 9 = (2r − 3)2 = 0,
3
3
which has one repeated root r = 23 . The fundamental solution set is {e 2 x , xe 2 x }
So, the general solution to the given equation is
3
3
3
y(x) = c1 e 2 x + c2 xe 2 x = (c1 + c2 x)e 2 x .
To find the solution to the initial value problem we have to plug x = 0 into y(x) and y ′(x).
3
3
3
3
3
3
y ′(x) = (c1 + c2 x)e 2 x + c2 e 2 x = ( c1 + c2 + c2 x)e 2 x ,
2
2
2
y(0) = c1 = 1,
Since c1 = 1 and c2 =
3
2
3
3
y ′ (0) = c1 + c2 = .
2
2
− 32 c1 = 0, the solution to the initial value problem is
3
y(x) = e 2 x .
Section 4.6 Auxiliary Equation with Complex Roots
If
√
b2 − 4ac < 0, then the equation (2) has two complex conjugate roots
√
4ac − b2
b
= α + iβ,
r1 = − + i
2a
2a
√
b
4ac − b2
r2 = − − i
= α − iβ = r¯1 ,
2a
2a
b
,β=
here i2 = −1, α = − 2a
√
4ac−b2
,
2a
α, β ∈ R.
We’d like to assert that the functions er1 x and er2 x are solutions to the equation (1). If we
assume that the law of exponents applies to complex numbers, then
e(α+iβ)x = eαx eiβx
eiβx −?
Let’s assume that the Maclaurin series for ez is the same for complex numbers z as it is for
real numbers.
iθ
e =
∞
X
(iθ)k
k=0
k!
= 1 + (iθ) +
(iθ)2
(iθ)k
+···+
+···
2!
k!
Since i2 = −1,
θ2
−
eiθ = 1 + (iθ) −
2!
θ2 θ4 θ6
1−
+
−
+··· +i θ −
2!
4!
6!
iθ3 θ4 iθ5
+
+
+··· =
3!
4!
5!
θ3 θ5 θ7
+
−
+ · · · = cos θ + i sin θ.
3!
5!
7!
So,
eiθ = cos θ + i sin θ,
then eiβx = cos βx + i sin βx and e(α+iβ)x = eαx eiβx = eαx (cos βx + i sin βx).
Lemma 1. Let z(x) = u(x) + iv(x) be a complex valued function of the real variable x,
here u(x) and v(x) are real valued functions. And let z(x) be a solution to the equation (1).
Then, the functions u(x) and v(x) are real-valued solutions to the equation (1).
Proof.
By assumption,
az ′′ + bz ′ + cz = a(u + iv)′′ + b(u + iv)′ + c(u + iv) =
a(u′′ + iv ′′ ) + b(u′ + iv ′ ) + c(u + iv) = (au′′ + bu′ + cu) + i(av ′′ + bv ′ + cv) = 0.
But a complex number a + ib = 0 if and only if a = 0 and b = 0. So,
au′′ + bu′ + cu = 0,
av ′′ + bv ′ + cv = 0,
which means that both u(x) and v(x) are real-valued solutions to (1).
When we apply Lemma 1 to the solution
e(α+iβ)x = eαx (cos βx + i sin βx),
we obtain the following.
Complex conjugate roots.
If the auxiliary equation has complex conjugate roots α ± iβ, then two linearly independent
solutions to (1) are eαx cos βx and eαx sin βx and a general solution is
y(x) = c1 eαx cos βx + c2 eαx sin βx,
where c1 and c2 are arbitrary constants.
Example 2. Find a general solutions.
(a) y ′′ − 10y ′ + 26y = 0.
SOLUTION
The associated auxiliary equation is
r 2 − 10r + 26 = 0,
which roots are r1 = −5+i, r2 = −5−i, the fundamental solution set is {e−5x cos x, e−5x sin x}.
Thus, the general solution is
y(x) = e−5x (c1 cos x + c2 sin x).
(b) y ′′ + 4y = 0.
SOLUTION
The associated auxiliary equation is
r 2 + 4 = 0,
which roots are r1 = 2i, r2 = −2i, the fundamental solution set is {cos 2x, sin 2x}. Thus,
the general solution is
y(x) = c1 cos 2x + c2 sin 2x.
(c) y ′′′ − y ′′ + 4y ′ − 4y = 0.
SOLUTION
The associated auxiliary equation is
r 3 − r 2 + 4r − 4 = r 2 (r − 1) + 4(r − 1) = (r 2 + 4)(r − 1) = 0,
which has one real root r1 = 1 and two complex roots r2 = 2i, r3 = −2i, so the fundamental
solution set is {ex , cos 2x, sin 2x}. Thus, the general solution is
y(x) = c1 ex + c2 cos 2x + c3 sin 2x.
Example 3. Solve the given initial value problems.
(a) w ′′ − 4w ′ + 5w = 0, w(0) = 1, w ′ (0) = 4.
SOLUTION The associated characteristic equation is
r 2 − 4r + 5 = 0,
which has two complex roots
r1 = 2 + i,
r2 = 2 − i.
So, the general solution to the given equation is
w(x) = e2x (c1 cos x + c2 sin x).
To find the solution to the initial value problem we have to plug x = 0 into w(x) and w ′(x).
w ′ (x) = 2e2x (c1 cos x + c2 sin x) + e2x (−c1 sin x + c2 cos x) = e2x ((2c1 + c2 ) cos x + (2c2 − c1 ) sin x),
w(0) = c1 = 1,
w ′(0) = 2c1 + c2 = 4.
Since c1 = 1 and c2 = 4 − 2c1 = 2, the solution to the initial value problem is
w(x) = e2x (cos x + 2 sin x).
(b) y ′′ + 9y = 0, y(0) = 1, y ′(0) = 1.
SOLUTION The associated characteristic equation is
r 2 + 9 = 0,
which has two complex roots
r1 = 3i,
r2 = −3i.
So, the general solution to the given equation is
y(x) = c1 cos 3x + c2 sin 3x.
To find the solution to the initial value problem we have to plug x = 0 into y(x) and y ′(x).
y ′ (x) = −3c1 sin 3x + 3c2 cos 3x,
y(0) = c1 = 1,
y ′ (0) = 3c2 = 1.
Since c1 = 1 and c2 = 1/3, the solution to the initial value problem is
y(x) = cos 3x +
1
sin x.
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